The Mathematics of Zome

AbstractWe will calculate the lengths of all the Zome struts and show that the locations of the Zome balls inan arbitrary configuration satisfy a very simple rule that involves the golden ratio.

1 IntroductionThe Zome system is a construction system based on a set of plastic struts and balls that can be attachedtogether to form an amazing set of mathematically or artistically interesting structures. There is a fairamount of deep mathematics involved, and the purpose of this article is to look at some of that.Anyone who has played with the Zome system is usually amazed at first how well things work out. Itallows one to construct a huge number of structures, but no matter how complex the structure, it seemsthat if two zome balls are near each other and each has a hole pointing at the other, the holes will be ofthe correct shape and in the correct orientation that a standard Zome strut will connect them. There is amathematical reason for this, and the purpose of this article is to demonstrate that reason. The article doesnot concern how to build complex Zome structures. For that sort of information, there are hundreds ofresources on the internet.It will be much easier to follow along if you have at least a small set of Zome parts (and the larger yourset, the better, although large sets tend to become expensive). It is also useful for you to have fooledaround a bit with the system before trying to understand some of the mathematics expressed here. Thisarticle includes photos of various Zome structures, but you will usually find the arguments much easierto follow if you build a physical model of each of those structures yourself so that you can more easilymanipulate the 3-D version and see it from many points of view.For information on Zome and for an on-line way to order kits or parts, see:http://www.zometool.comThe main Zome strut colors are red, yellow and blue and most of what well cover here will use those asexamples. There are green struts that are necessary for building structures with regular tetrahedrons andoctahedrons, and almost everything we say about the red, yellow and blue struts will apply to the greenones. The green ones are a little harder to work with (both physically and mathematically) because theyhave a pentagon-shaped head, but can fit into any pentagonal hole in five different orientations. With theregular red, yellow and blue struts there is only one way to insert a strut into a Zome ball hole. The bluegreen struts are not really part of the Zome geometry (because they have the wrong length). They arenecessary for building a few of the Archimedian solids, like the rhombicuboctahedron and the truncatedcuboctahedron.

2 The Zome Ball

Look carefully at a Zome ball. (It is better to look at a physical ball, but an image of one appears inFigure 1. It is highly symmetric, and has holes that will accept struts of three shapes: rectangles with an

aspect ratio of roughly 1 : 2, equilateral triangles and regular pentagons. Every pentagonal hole looks thesame: it is surrounded by 5 rectangular holes and 5 triangular holes. The same can be said of every hole:the shapes and orientations of the neighboring holes are the same for every hole in the ball.

Figure 1: The Zome Ball

Another way to convince yourself that all the holes of a certain shape are basically identical is to place aZome ball on a table balanced on a hole of a particular shape (say a rectangle). Now take another Zomeball and place it on any of its rectangular holes (or hole of the same shape as the first ball). If you rotate thesecond ball so that the rectangles on top have the same orientation, you will find that every hole matchesin shape and direction in the two balls.There are 12 pentagon-shaped holes and if you imagine that the pentagons were all left in their planes butexpanded until their edges touched the nearest pentagon edges, the resulting figure would be a dodecahedron (a regular 12-sided polyhedron).If you think about this pentagon expansion, every pair of adjacent pentagons would close over a rectangular hole, so there are the same number of rectangular holes as there are edges in a dodecahedron; namely,30.Finally, again visualizing the expansion of the pentagonal holes, each triangle on the Zome ball will becovered by a vertex of the final dodecahedron, so there are the same number of triangles as there arevertices of a dodecahedron; namely, 20.A dual argument can be made: instead of expanding the pentagons until their edges merge, expand thetriangles in the same way, and the resulting figure will be a regular icosahedron a polyhedron with 20identical triangluar sides. Each vertex of the resulting icosahedron (of which there are 12) corresponds toa pentagonal hole in the Zome ball and each edge of the icosahedron (of which there are 30) correspondsto one of the rectangular holes in the Zome ball.Luckily, we obtain the same counts using both approaches: 12 pentagonal holes, 20 triangular holes and30 rectangular holes for a total of 12 + 20 + 30 = 62 holes.Finally, lets look at the orientations of Zome balls in any sort of structure. If you hook together any sortof combinations of balls and struts in such a way that none of the struts are forced to bend, then everyZome ball in the entire structure will have exactly the same orientation: if one ball has a rectangle pointingin a particular direction with a particular orientation, then every ball in the structure will have one of itsrectangles pointing in the same direction and with the same orientation. Figure 2 shows a central ball withone strut of each of the major colors and with a Zome ball attached to the end of each. Note that the ballsall have the same orientation.It is pretty easy to convince yourself of this: just place a strut of each shape into a Zome ball and look atthe balls on the other ends of the struts. In every case, the adjacent ball has the same orientation, so if youbegin at any ball and follow a sequence of struts to another one, youll wind up at a ball that has the sameorientation. This is true even of the green (and even of the blue-green) struts.

Figure 2: Zome Ball Orientation

3 Zome StrutsIn a standard Zome set there are struts of three lengths in each color: red, yellow and blue. It is nowpossible to purchase other lengths of some of them: short and super-short. We will call the original ninestruts the standard struts.The usual labeling of the standard struts is B1 , B2 , B3 , Y1 , Y2 , Y3 , R1 , R2 and R3 for the blue, yellowand red struts, and the smaller numbers refer to the shorter struts. Thus B1 is the shortest blue strut andR3 is the longest red one.The physical lengths of the struts are chosen so that their mathematical lengths are perfect. If we considera structure that has a Zome ball at the end of every strut, in a mathematical sense, the centers of theZome balls should be considered to be the endpoints of the struts embedded in the balls. Thus the truemathematical length of a strut should be the distance between the centers of two Zome balls attached tothe ends of the strut.Measuring in terms of these mathematical lengths, the strut lengths follow a very regular pattern: in eachcolor, the lengths increase by a constant factor called (this is the Greek letter tau and it stands for theso-called golden ratio: a number that is approximately 1.618).Thus the mathematical length of B2 is times as long as a B1 , and the length of B3 is times the lengthof B2 (or 2 times the length of B1 ). The same relation holds for struts of every color, including thegreens and even the blue-greens.We will assign an arbitrary length of 1 to B1 , the shortest blue strut. With this assignment, the lengths ofB2 and B3 are therefore and 2 , respectively.From Figure 3 we can glean enough information to calculate the lengths of the yellow struts. The planarfigure is made of two perpendicular blue struts, B1 and B3 . The three yellow struts all are of type Y2 . Itis clear from the construction that a yellow Y2 strut is the hypotenuse of a right triangle whose sides havelengths 1/2 and 2 /2. Using the Pythagorean theorem and a bit of messy algebra (which we will be ableto do much more easily after we haveexamined some properties of in Section 4.1)yields the result thatthe length ofthe Y2 yellow strut is ( 3/2). This implies that the length of Y1 is ( 3/2) and the lengthof Y3 is 2 ( 3/2) since all the struts of the same color have lengths that are simply multiples of of eachother.

Figure 3: Yellow Struts

Figure 4: Red Struts

Similarly, the length of the red struts can be calculated from the planar Figure 4. This time the structure isbuilt from two perpendicular blue struts B1 and B2 and the red struts are all R1 . This time the R1 formsthe hypotenuse of a right triangle whose other sides have lengths 1/2 and /2. Again, an ugly calculation

yields the length of R1 as 2 + /2, so the lengths of R2 and R3 are 2 + /2 and 2 2 + /2,respectively. See Section 4.1 for the details of the calculation.

4 The Golden Ratio

The number is defined to be the largest root of the equationx2 = x + 1.

(1)

which can easily be solved using the quadratic formula or any other method, and the resulting value for is (1 + 5)/2. But equation 1 allows us to express powers of in terms of itself. Obviously we obtaindirectly from equation 1 that 2 = + 1.What about 3 ? Well, 3 = ( 2 ) and we know that 2 = + 1, so 3 = ( + 1) = 2 + = 2 + 1.We can, using the same approach, obtain expressions for 4 , 5 , and so on, yielding the following table:

123456

======

112358

+0+1+1+2+3+5

A quick glance at the table shows that the coefficients of the expansion are just the Fibonacci numbers,and a simple inductive argument shows us that this is true.Since the lengths of the Zome struts of any color are just multiples of a basic length by some power of ,then every strut length can also be expressed as an integer linear combination of 1 and times the lengthof the basic strut of that color.A standard Zome set has a certain shortest length of the blue strut which we defined to be 1, but in principle, shorter struts could be obtained by continuing to divide the lengths by . It is somewhat amazing,but dividing by powers of also yields integer combinations of 1 and . For example,1= 1,

and this can be obtained from equation 1 by dividing both sides by and regrouping terms. But now thatthe value of 1/ is known, we can obtain in a similar way values for 1/ 2 , 1/ 3 , et cetera, and the tableabove can be extended in the negative direction. It is a good exercise for the reader to check these values: 4 3 2 101234

=========

3 + 52 31 + 21 10 + 11 + 01 + 12 + 13 + 2

The coefficients for the negative powers of look a little strange: the signs alternate, but the values arejust the familiar Fibonacci numbers. The usual definiton of the Fibonacci numbers starts with F0 = 0 andF1 = 1, and once any two adjacent values are known, the next is obtained using the recurrence relationFn = Fn1 + Fn2 .The nice thing is that the same recurrence relation can be used to obtain reasonable values for F1 , F2and so on. What should be the value, for example, of F1 ? Well, it should satisfy:F1 + F0 = F1 ,and since we know the values of F0 and F1 , we obtain F1 = 1.Continuing in the same way, we can obtain F2 = 1, F3 = 2, F4 = 3 and so on. Note that thesefollow the same pattern as the coefficients of the simplified negative powers of in the table above. It iseasy to show that this pattern continues. In fact, the general formula for n , where n is positive, negative,or zero, is given by: n = Fn + Fn+1 ,where the values of the Fibonacci numbers are extended to negative values as described above.

4.1 Calculating Strut Lengths

Using the relations we have discovered in the first part of this section, it is easy to work out the lengths ofthe red and yellow struts in terms of the blue ones.5

Lets begin with the yellow struts. As we saw in Section 2, the length of Y2 is the hypotenuse of a righttriangle with sides having lengths 1/2 and 2 /2. If that unkown length is y2 , we have:r 1 2 2 2+.y2 =22Thus:y22 =

1 + 4.4

Since 2 = 1 + , we can square both sides to obtain 4 = (1 + )2 which we can substitute into theequation above:y22

which leads to the values for the lengths of Y1 , Y2 and Y3 : 3/2, 3/2, and 2 32, respectively.y22

To calculate the length of the red struts, we again use the Pythagorean theorem in conjunction with thestructure illustrated in Figure 4. The length r1 of R1 is given by:r 1 2 2r1 =+2221 +r12 =441 + 2=.4Since 2 = 1 + , this becomes:2+42+.r1 =2

Thus the lengths of R2 and R3 are 2 + /2 and 2 2 + /2, respectively.

r12

5 Zome Ball Coordinates

Because of the angles of the Zome ball holes, it turns out that any combination of red, blue, yellow (andeven green) struts will leave a ball at a point with nice coordinates, based on the following coordinatesystem:

As before, call the length of the shortest blue strut 1. Plug three of these shortest blues into a Zome ball sothat they are all mutually perpendicular, and form a right-handed coordinate system. If that original Zomeball has coordinates (0, 0, 0), then the centers of Zome balls stuck on the ends of each of the coordinateaxis struts will have coordinates (1, 0, 0), (0, 1, 0) and (0, 0, 1). This will be our coordinate system and wewill now examine the coordinates of the centers of Zome balls which are hooked together in an arbitraryfashion to a ball that is designated to lie at the origin of our system with some set of short blues markingthe x, y and z axes.It turns out that no matter what combinations of struts of any lengths are stuffed into the holes, with anylinkage whatsoever, the coordinates of any reachable Zome ball will have the form (, , ), where , and are numbers of the form (a + b)/2, where a and b are integers (possibly negative or zero) and isthe golden ratio: = (1 + 5)/2 1.618039887.

Now we will demonstrate our main result, that the coordinates of any zome ball reachable from a Zomeball at (0, 0, 0) all have the form:a + b c + d e + f(,,)(2)222What we need to do to prove this is to show that when any strut is plugged into any Zome ball holehaving coordinates like those above, the coordinates of a ball placed at the end of the strut will alsohave coordinates with the same property. Since the ball at the origin satisfies the condition (namelya = b = c = d = e = f = 0) then the condition is preserved as each strut is added to any spatial path.Note: It is possible to purchase special half blue and half green struts that are half the length of thenormal ones. If these are allowed in the system, the result holds except that the denominator in Equation 2must be changed from 2 to 4.This is true of the red, blue, yellow and green struts, but we will only prove it here for the red, blue andyellow ones. The proof for the green struts is similar, but there are five possibilities to consider since agreen strut can be pushed into a pentagonal hole in five different orientations. Working out the details forthe green struts is not difficult, but there are five cases, so it is a good exercise to do this.

Figure 5: Yellow Strut

We will demonstrate our result for the red, blue and yellow struts by showing that if a particular strutof a particular color is put in the Zome ball at the origin, the coordinates of the center of the ball atthe end of the strut have the correct form. Then, since the actual strut used can have a length that is amultiple by a power of of the particular strut, the coordinates for a different length strut will simply

be multiplied by some power of . But we have shown that such a multiplication will just yield anotherinteger combination of 1 and , so every strut coming out of the origin will yield an endpoint having theform shown in equation 2.Finally, since we will be, in general, placing the strut into a ball that is not at the origin, but into a ballwhose coordinates have the form shown in equation 2, we will simply be adding coordinates of thoseforms, and the resulting ball at the end of the strut will have the required coordinates.Insert three short blue struts to represent the perpendicular coordinate axes and then add three moreopposite them to represent the negative axes. When you examine the Zome ball with these six positiveand negative axes, they divide the ball into eight octants that, apart from sign and labeling of the axes,are equivalent. Thus we really need only examine the holes in the Zome ball that fall into one octant orboundary of that octant. A quick look will make it obvious that in each octant there is essentially only twokinds of holes into which a blue strut can be inserted (an axis and a non-axis blue hole), two into whicha yellow can be inserted (in one of the planes determined by two axes and one making equal angles withall the axes, and only one for the red struts (again in a plane determined by two of the axes).We need to show that if we assume that the Zome ball has coordinates (0, 0, 0) the other end of each ofthose five types of struts will have coordinates satisfying Equation 2.

Figure 6: Blue Struts

First note that if any particular strut satisfies these conditions, then all struts of the same color insertedinto the same hole will work, since every struts length is a power of times another strut of the samecolor, and we know that multiplying coordinates by a power of can always be reduced to a value that islinear in 1 and as we showed in Section 4.The easiest struts to consider are the blue axial struts. With a short blue, the coordinates will always looksomething like (1, 0, 0), (0, 1, 0) or (0, 0, 1), and these all satisfy the conditions of equation 2.

The red and yellow struts that lie in a plane determined by a pair of axes are also quite easy. In fact,Figure 3 shows that a Y2 strut from the origin has coordinates 1/2, 2 /2 = (1 + )/2, and 0 in someorder, depending on the particular pair of axes that determine the planeFigure 4 gives a similar argument that the R1 struts have coordinates 1/2, and 0, in some order.

A yellow strut that makes equal angles with all three axes is shown in Figure 5. If you construct it withthree B1 axes and three Y1 struts as shown, it is clear that the ball in the center of the yellow struts hascoordinates (1/2, 1/2, 1/2).8

Finally, the trickiest standard strut turns out to be a blue strut in one of the off-axis holes.In Figure 6, the coordinate axes are made of a B1 , a B2 and a B3 . From the illustration, it is clear thatthe central ball, relative to the origin, has coordinates 1/2, /2, and 2 /2 = (1 + )/2, in someorder.An interesting consequence of this is that if we had blue struts in every length, including 1/ , 1/ 2 , . . . ,and if all we cared about were the locations of the Zome balls in the final structure, then there would beno need for the yellow, red or green balls since simple movements in the directions of the coordinate axeswith these blue struts can get us to any ball location reachable using the red, yellow and green struts.